2014微软实习生招聘第二题

Time Limit: 10000ms

Case Time Limit: 1000ms

Memory Limit: 256MB

Description

Consider a string set that each of them consists of {0, 1} only. All strings in the set have the same number of 0s and 1s. Write a program to find and output the K-th string according to the dictionary order. If s​uch a string doesn’t exist, or the input is
not valid, please output “Impossible”. For example, if we have two ‘0’s and two ‘1’s, we will have a set with 6 different strings, {0011, 0101, 0110, 1001, 1010, 1100}, and the 4th string is 1001.

Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 10000), the number of test cases, followed by the input data for each test case.

Each test case is 3 integers separated by blank space: N, M(2 <= N + M <= 33 and N , M >= 0), K(1 <= K <= 1000000000). N stands for the number of ‘0’s, M stands for the number of ‘1’s, and K stands for the K-th of string in the set that needs to be printed
as output.

Output

For each case, print exactly one line. If the string exists, please print it, otherwise print “Impossible”.

Sample In

3

2 2 2

2 2 7

4 7 47

Sample Out

0101

Impossible

01010111011

我的细节处理并不高明，但是套路和大家都相似，对第k个字典序string temp，从后往前遇到'1','0'可交换时候，把此前的'1'全部放到最后。直接上代码

#include "stdafx.h"
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>
using namespace std;

bool permutatiom(string s, int k, string &result,int n,int m)
{
string temp=s;
string out;
for(int i=0;i<m;i++)
{
out.push_back('1');
}
for(int i=0;i<n;i++)
{
out.push_back('0');
}
int i=1;
for( ;i<k && temp.compare(out)!=0;i++)
{
int count=0;
for(int j=n+m-1;j>0;j--)
{
//int count=0;
if(temp[j]=='1')
count++;
if(temp[j]=='1' && temp[j-1]=='0')
{
temp[j]='0';
temp[j-1]='1';
count--;
for(int ii=m+n-1;ii>m+n-1 -count;ii--)
{
temp[ii]='1';
}
for(int ii=m+n-1-count;ii>j;ii--)
{
temp[ii]='0';
}
break;
}
}
//cout<<temp<<endl;
}
if(i==k)
{
result=temp;
return true;
}
else
{
//cout<<i<<endl;
//cout<<temp<<endl;
return false;
}
}

int main()

{
int ss;
int n,m,k;
cin>>ss;
while(ss--)
{
cin>>n>>m>>k;
if(k<=0)
{
cout<<"Impossible"<<endl;
break;
}
string s;
for(int i=0;i<n;i++)
{
s.push_back('0');
}
for(int i=0;i<m;i++)
{
s.push_back('1');
}
string result;
if(permutatiom(s,k,result,n,m))
{
cout<<result<<endl;
}
else
cout<<"Impossible"<<endl;

}
//system("pause");
return 0;
}